September 30, 2012

We consider the question of when a perfect square is equal to the difference of two consecutive cubes. If we let the cubes be and , we see that we require to be a perfect square, say .

If you list the first cubes they are and their differences are . Of these the squares are and . By continuing to list cubes it would take a while to find the next instance of a square as a difference – the next solution is . 🙂

Initially I tried solving this problem by rearranging as but that did not take me far. Another approach was via congruences, noting that the right side is so that had to be of the form . The third and most fruitful approach for me was to view the equation as a quadratic in .

Since this has integer roots, its discriminant must be a perfect square. In other words, for some integer . This equation implies must be a multiple of 3, so letting we obtain , or

In other words, we need to find (positive) integer solutions to where is even. An equation of this form is known as Pell’s equation. We claim that solutions to this equation are of the form , where is the smallest positive solution to while .

To see why this is so, firstly note that if is another positive solution with , then the equation leads to , so and . Hence it makes sense to define the “smallest solution”.

Next, we know there exists some for which

.

Multiplying all sides by and using the fact that gives us

Define . We can use norms in or explicit computation to show that . Equation (2) then contradicts the minimality assumption on unless we have . We conclude that , so all solutions are of this form proving the claim.

In our case, the minimal solution is , so we have the general solution

We can use the recurrence to see that and have the same parity (odd or even), as do and . Since initially , we see that the solutions alternate in parity. Since we are looking for being even, we are looking for odd .

.

Using the same method as before this leads to the recurrences

We can decouple these by taking 7 times the first equation minus 12 times the second to obtain . Then . Similarly we find .

Initial solutions are . From (1) we have , so we require to be odd. This will indeed be the case by our parity arguments from before (that and have different parity). We conclude that this recurrence gives us the entire solution set for . The non-negative solutions are . In other words,

The recurrence linking these can be found to be and and it can be verified that . The factor of 14 in the recurrence indicates that the solution set is relatively sparse (though still infinite).

A similar approach is shown in [3], where the Pell’s equation is obtained by completing the square to obtain rather than using the discrminant.

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September 27, 2012

Recently Hashim Amla from South Africa performed a rare feat in averaging over 100 in both test and ODI cricket series against England. It led me to look up others who have performed similarly. The following lists in chronological order those who have averaged over 100 in both test and ODI series against a single team with the restriction that at least 180 ODI runs were scored. I collected the data manually starting with this list of high ODI series averages. Corrections/additions are welcome.

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September 23, 2012

It has only 3 different digits (consecutive ones at that and including 6 and 7).

The first three digits of the number are the same as the first three digits of the original expression . They are also easy to recall if you link it to when the ancient Olympics were first held or when the US Declaration of Independence was signed. (I often link numbers to years of events as a memory aid.)

The two three-digit numbers 776 and 887 making up the six-digit number differ by 111. 🙂

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September 13, 2012

This week Andy Murray broke through with his first grand slam singles title in the 2012 US Open. This led me to compile the following list of the grand slam singles champions (winning for the first time in the Open era from 1968) along with the number of tournaments they won before winning that first grand slam. As can be seen, only a handful won more than a dozen tournaments before their first major, with Ivan Lendl leading the way with 40 tournament wins spanning four years before his first major! Aside from Murray and his coach Lendl, Chris Evert, Martina Navratilova, Jana Novotna, Kim Clijsters, Guillermo Vilas, Thomas Muster and Goran Ivanišević won more than 20 tournaments before finally achieving that first grand slam win.

A few players won a grand slam as their first tournament victory, Gustavo Kuerten the most recent to do so. As far as I can tell only Chris O’Neil won her one and only singles tournament at a grand slam.

As an aside, I noticed that Thomas Muster had an amazing run in 1995, winning 11 tournaments in just over six months, all on clay!